28
Oct 17

## The Lagrangian multiplier interpretation

Motivation. Consider the problem of utility maximization $\max u(x,y)$ under the budget constraint $p_xx+p_yy=M.$ Suppose the budget $M$ changes a little bit and let us see how the solution of the problem depends on changes in the budget. The maximized value of the utility will depend on $M$, so let us reflect this dependence in the notation $U(M)=u(x^\ast(M),y^\ast(M))$ where, for each $M$, $(x^\ast(M),y^\ast(M))$ is the maximizing bundle. The value of the Lagrangian multiplier $\lambda$ in the FOC's $\frac{\partial L}{\partial x}=0$, $\frac{\partial L}{\partial y}=0$, $\frac{\partial L}{\partial \lambda }=0$ will also depend on $M$: $\lambda =\lambda(M).$ The property that we will prove is

$\frac{dU}{dM}=\lambda (M),$

that is, the Lagrangian multiplier measures the sensitivity of the maximized utility function to changes in the budget.

## General formulation

The main problem is to maximize $f(x,y)$ subject to $g(x,y)=0$. However, instead of a fixed constraint we consider a set of constraints perturbed by a constant $c$: $g(x,y)+c=0.$ Here $c$ varies in a small neighborhood $(-\varepsilon ,\varepsilon)$ of zero (which corresponds to varying $M$ in a neighborhood of some $M_{0}$ in the motivating example). Now everything depends on $c$, and differentiation with respect to $c$ will give the desired result.

## Employing the constraint

As before, we assume the implicit function existence condition:

$\left( \frac{\partial g}{\partial x},\frac{\partial g}{\partial y}\right)\neq 0.$

Changing the notation, if necessary, we can think that it is the last component of the gradient that is not zero: $\frac{\partial g}{\partial y}\neq 0.$ This condition applies to the perturbed constraint too and guarantees existence of the implicit function, which now depends also on $c$: $y=y(x,c)$. Plugging it into the constraint we get $g(x,y(x,c))+c=0$, and differentiation yields

(1) $\frac{\partial g}{\partial y}\frac{\partial y}{\partial c}+1=0.$

## Employing the FOC's

Critical assumption. For each $c\in (-\varepsilon ,\varepsilon )$ the perturbed maximization problem has a solution. At least in our motivating example, this assumption is satisfied.

The maximized objective function is denoted $F(c)=f(x,y(x,c)).$ At each $c,$ we have the right to use the FOC's for the Lagrangian. One of the FOC's is

$\frac{\partial f}{\partial y}+\lambda (c)\frac{\partial g}{\partial y}=0.$

To make use of (1), multiply this by $\frac{\partial y}{\partial c}$:

$0=\frac{\partial f}{\partial y}\frac{\partial y}{\partial c}+\lambda(c)\frac{\partial g}{\partial y}\frac{\partial y}{\partial c}=\frac{\partial f}{\partial y}\frac{\partial y}{\partial c}-\lambda (c).$

It follows that

$\frac{dF}{dc}=\frac{\partial f}{\partial y}\frac{\partial y}{\partial c}=\lambda (c).$

With $c=0$ we have $\lambda(c)=\lambda$, the Lagrange multiplier for the unperturbed problem, and as a result

$\frac{dF}{dc}=\frac{\partial f}{\partial y}\frac{\partial y}{\partial c}=\lambda.$